Question

Find the remainder when ${{x}^{3}}+3{{x}^{2}}+3x+1$ is divided by $x+1$

Answer 1

Hint: We solve this problem by using the long division method. We divide the given polynomial with the polynomial $x+1$ and find the remainder.
In dividing one polynomial with another polynomial we take the quotient by dividing the first term of dividend with the first term of divisor and take that in quotient. Then we multiply the whole divisor with that quotient and subtract from the dividend to get the remainder. We carry out this process until the degree of remainder is less than that of the divisor.

Complete step by step solution:
We are asked to find the remainder when ${{x}^{3}}+3{{x}^{2}}+3x+1$ is divided by $x+1$.
Here, we can see that the polynomial ${{x}^{3}}+3{{x}^{2}}+3x+1$ is getting divided so that it will be the dividend.
Also we can see that the polynomial $x+1$ is used to divide so it is the divisor.
Now, let us carry out the long division process then we get,
$x+1\overline{\left){{{x}^{3}}+3{{x}^{2}}+3x+1}\right.}$
Now, we know that the quotient is obtained by dividing the first term of dividend with the first term of divisor.
Here, we can see that the first term of dividend is ${{x}^{3}}$ and first term of divisor is $x$ which leaves the quotient as ${{x}^{2}}$
Now, by taking ${{x}^{2}}$ as the quotient in the above division and subtracting the polynomial we get after multiplying the whole divisor with the quotient we take then we get,
$\begin{align}
  & x+1\overset{{{x}^{2}}}{\overline{\left){{{x}^{3}}+3{{x}^{2}}+3x+1}\right.}} \\
 & \text{ }\underline{-\left( {{x}^{3}}+{{x}^{2}} \right)\text{ }} \\
 & \text{ }2{{x}^{2}}+3x+1 \\
\end{align}$
Here, we can see that the degree of remainder is 2 and the degree of divisor is 1 in which the division is not yet completed.
Here, we can see that the first term of remainder is $2{{x}^{2}}$ and first term of divisor is $x$ which leaves the quotient as $2x$
Now, by taking $2x$ as the quotient in the above division and subtracting the polynomial we get after multiplying the whole divisor with the quotient we take then we get,
$\begin{align}
  & x+1\overset{{{x}^{2}}+2x}{\overline{\left){{{x}^{3}}+3{{x}^{2}}+3x+1}\right.}} \\
 & \text{ }\underline{-\left( {{x}^{3}}+{{x}^{2}} \right)\text{ }} \\
 & \text{ }2{{x}^{2}}+3x+1 \\
 & \text{ }\underline{-\left( 2{{x}^{2}}+2x \right)\text{ }} \\
 & \text{ }x+1 \\
\end{align}$
Here, we can see that the degree of remainder is 1 and the degree of divisor is 1 in which the division is not yet completed.
Here, we can see that the first term of remainder is $x$ and first term of divisor is $x$ which leaves the quotient as $1$
Now, by taking $1$ as the quotient in the above division and subtracting the polynomial we get after multiplying the whole divisor with the quotient we take then we get,
$\begin{align}
  & x+1\overset{{{x}^{2}}+2x+1}{\overline{\left){{{x}^{3}}+3{{x}^{2}}+3x+1}\right.}} \\
 & \text{ }\underline{-\left( {{x}^{3}}+{{x}^{2}} \right)\text{ }} \\
 & \text{ }2{{x}^{2}}+3x+1 \\
 & \text{ }\underline{-\left( 2{{x}^{2}}+2x \right)\text{ }} \\
 & \text{ }x+1 \\
 & \text{ }\underline{-\left( x+1 \right)\text{ }} \\
 & \text{ }0 \\
\end{align}$
Here, we can see that the degree of remainder is 0 which is less than that of the divisor which says that the division is complete.
Therefore, we can conclude that the remainder when ${{x}^{3}}+3{{x}^{2}}+3x+1$ is divided by $x+1$ is ‘0’.

Note: We have a shortcut for solving this problem when the divisor is linear polynomial.
The remainder theorem states that if any polynomial $p\left( x \right)$ is divided by a linear polynomial $x-a$ then the remainder is given as,
$R=p\left( a \right)$
Now, let us use this remainder theorem for the given polynomials.
We are given that ${{x}^{3}}+3{{x}^{2}}+3x+1$ is divided by $x+1$
By comparing the given polynomials with the polynomials in remainder theorem then we get,
$\Rightarrow p\left( x \right)={{x}^{3}}+3{{x}^{2}}+3x+1$
Also, the value of $'a'$ is given as,
$\Rightarrow a=-1$
Now, by using the remainder theorem we get the required remainder as,
$\begin{align}
  & \Rightarrow p\left( -1 \right)={{\left( -1 \right)}^{3}}+3{{\left( -1 \right)}^{2}}+3\left( -1 \right)+1 \\
 & \Rightarrow p\left( -1 \right)=-1+3-3+1 \\
 & \Rightarrow p\left( -1 \right)=0 \\
\end{align}$
Therefore, we can conclude that the remainder when ${{x}^{3}}+3{{x}^{2}}+3x+1$ is divided by $x+1$ is ‘0’.